Question

The time taken by the milk to sour at ${37}^{\circ }C$

• A. $35.2$ hr
• B. $32.5$ hr
• C. $35.3$ hr
• D. $32.3$ hr

Answer 1

The correct option is B $32.5$ hr

Now $T_1=300K$, $T_2={37}^{\circ }C=310K$

$E_a=30146Jmo{l}^{-1}$

log $\frac{k_{310}}{k_{300}}=\frac{30146Jmo{l}^{-1}}{2.303\times 8.314}\times \frac{310-300}{300\times 310}=0.6193$

$\therefore$ $\frac{k_{3}10}{k_{3}00}=antilog\left(0.1693\right)=1.477$

Higher the rate constant, faster is the reaction, i.e., lesser is the time taken, hence

$\frac{t_{3}10}{t_{3}00}=\frac{k_{3}00}{k_{3}10}=\frac{I}{1.477}$

$\therefore t_{3}10=t_{3}00\times \frac{1}{1.477}=\frac{48}{1.477}=32.5hr$